00:01
You want to find all distinct left co -sets of h, h being this set 1 and 11.
00:18
And so to find all the left co -sets, you're going to multiply on the left mod 30 by each element.
00:28
So if i do 1 times h, of course, that's just going to be 1 and 11.
00:37
Now i'm going to do 7 times h, which 7 times.
00:42
Times one of course is seven.
00:45
Now 11 times seven is 77, but here i have a modulator.
00:50
This is a big help.
00:52
Because if i go to this, 77 mod 30, i can see that that's 17.
00:58
And this will really speed you up if you use this tool.
01:02
So 7 and 17 is 7 times h.
01:11
11 times h is going to be 1.
01:19
And 11 times 11 is 121 so that should just be one check check check so hopefully that i'm anyway what i was saying is that 11 times age will give you 11 and 121 and 121 mod 30 is 1 so then you just get back to 11 11 which shouldn't be a surprise because 11 is a member of age anyway and and 13.
01:56
So 13 times age.
01:57
Notice i'm multiplying each of these on the left because you want left co -sets.
02:04
That's gonna give us 13 and what is 13 times 11? 13 times 11 is, of course i know how to do that, but let's be us up.
02:23
So 13 times 11 is 143.
02:27
So what's 143, mod 30? 23, so 13 and 23.
02:35
And we keep going in this manner.
02:43
I'm going to stop the video here and calculate the rest of these.
02:47
All right and we're back.
02:50
Now if you want all the distinct left co -sets, notice that they're going to be 111.
03:04
You got a 717 in there.
03:11
You got a 1323 in there, and you got a 1929 in there.
03:31
And so there are four distinct left coasts sets of u -30 in that case.
03:39
Now the question does left equal right? that should be easy to answer because after all you're, excuse me, after all you're in an abelian group, u -30 with multiplication, the multiplication is commutative...